Equation (1) represents the membrane dynamics of the neuron \(i\) . \(C\) 
(\(\mu \) F/cm\(^2\) ) is the membrane capacitance and \(I_i\)  (\(\mu \) A/cm\(^2\) ) is a
constant current density chosen in the interval \([10,11]\) . The parameters
\(g_{\rm K}\) , \(g_{\rm Na}\) , and \(g_{\rm L}\)  are the conductance of the potassium,
sodium, and leak ion channels, respectively. \(E_{\rm K}, E_{\rm Na}\) , and
\(E_{\rm L}\)  are the reversal potentials for these ion channels. \(V_{\rm r}^{\rm +}\) 
corresponds to the excitatory reversal potential. \(g_{ij}\)  is the excitatory
coupling strength from the presynaptic neuron \(j\)  to the postsynaptic neuron
\(i\)  with maximum and minimum value within the interval \([0.0,0.01]\) . We consider
that the neuron has no self-connections, implying \(g_{ii}=0\) . \(f_i(t)\)  is
the normalised synaptic current from the neuron \(j\)  to \(i\) . The state variable
\(f_i\)  decays exponentially and it is updated to the unity (\(f_i\rightarrow 1\) )
at the spike time \(t_i\)  of the neuron \(i\)  [1]}, [2]}. Equation (5)
corresponds to an exponential decay on the evolution of \(f_i(t)\) . The parameter
\(\tau _{\rm s}\)  is the synaptic time decay and \(\tau _{ij}\)  the delay on the
synaptic transmission [3]}, [4]}, [5]}. The intensity of
synaptic current with time delay on the signal transmission depends
on the state of the pre and postsynaptic neuron. \(\tau _{ij}\)  assumes values
\(\tau _{\rm int}\)  and \(\tau _{\rm ext}\)  for internal and external connections between
the subnetworks, respectively. In Eqs. (2) and (3), the functions \(m(v_i)\)  and
\(n(v_i)\)  represent the sodium and potassium activation, respectively. In
Eq. (4), \(h(v_i)\)  is the function for sodium inactivation. The functions
\(\alpha _n\) , \(\beta _n\) , \(\alpha _m\) , \(\beta _m\) , \(\alpha _h\) , and \(\beta _n\)  are
given by
\(\alpha _n(v) & = & \frac{0.01v+0.55}{1-\exp \left(-0.1v-5.5\right)},\\\beta _n(v) & = & 0.125\exp \left(\frac{-v-65}{80}\right),\\\alpha _m(v) & = & \frac{0.1 v+4}{1-\exp \left(-0.1v-4\right)},\\\beta _m(v) & = & 4\exp \left(\frac{-v-65}{18}\right),\\\alpha _h(v) & = & 0.07\exp \left(\frac{-v-65}{20}\right),\\\beta _h(v) & = & \frac{1}{1+\exp \left(-0.1v-3.5\right)},\) 
